The equation of the curve passing through the origin and satisfying the differential equation $(1 + x^2) \frac{dy}{dx} + 2xy = 4x^2$ is

Let $y=y(x)$ be the solution of the differential equation $(x+1) y^{\prime}-y=e^{3 x}(x+1)^{2}$,with $y(0)=\frac{1}{3}$. Then,the point $x=-\frac{4}{3}$ for the curve $y = y ( x )$ is

If $y = y(x)$,$x \in \left(0, \frac{\pi}{2}\right)$ is the solution curve of the differential equation $(\sin^2 2x) \frac{dy}{dx} + (8 \sin^2 2x + 2 \sin 4x) y = 2 e^{-4x} (2 \sin 2x + \cos 2x)$,with $y\left(\frac{\pi}{4}\right) = e^{-\pi}$,then $y\left(\frac{\pi}{6}\right)$ is equal to:

If $y=y(x)$ is the solution of the differential equation $x \frac{dy}{dx} + 2y = x^2$ satisfying $y(1) = 1$,then the value of $y\left(\frac{1}{2}\right)$ is

The integrating factor of the differential equation $(1 + t^2) + (x - e^{\tan^{-1} t}) \frac{dt}{dx} = 0$ is

$A$ family of curves has the differential equation $x y \frac{d y}{d x}=2 y^2-x^2$. Then, the family of curves is